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Optimisation of a cylindrical container

  1. Jun 27, 2006 #1
    Hey
    I am given the set volume for a cylindrical container and separating pricing for the material used to construct the container, the base and the side wall. The ends of the container cost $0.05 per cm2 and the side walls $0.04 cm2, and the volume is 400 mL. The question asks: What is the best design for the container i.e., measurements to ensure that cost is at a minimum.

    _________________________________

    my working:
    V = 400 mL
    Ends costs $0.05 per cm2
    Side wall costs $0.04 cm2
    Area of ends = [tex]2 \pi r^2[/tex]
    Area of side wall [tex]2 \pi r h[/tex]
    V = 400, therefore since V = [tex]\pi r^2 h[/tex], [tex]h = \frac{400}{\pi r^2}[/tex] (1)
    Cost for one container = [tex]0.05 (2 \pi r^2) + 0.04 (2 \pi r h)[/tex] (2)
    Therefore substituting (1) into (2), cost = [tex]0.05 (2 \pi r^2) + 0.04 (2 \pi r \frac{400}{\pi r^2})[/tex]
    Differentiating the cost function gives: [tex]\frac {dC}{dr} = 0.2 \pi r + \frac {96}{r^2}[/tex]
    The value of [tex]r[/tex] where there is no rate of change = -5.34601847

    _________________________________

    I am not sure if what I have done after and including the differentiation is correct. Since I have got a negative answer for the value of r, I am thinking that what I have done is wrong, as you cannot have a negative radius. Would the second derivative help with the sloving this equation, if so how?
    Many thanks,
    Pavadrin
     
  2. jcsd
  3. Jun 27, 2006 #2

    HallsofIvy

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    Go ahead and write this out: cost= [tex]0.1 \pi r^2+ \frac{32}{r}[/itex]
    Not quite. It gives [tex]\frac{dC}{dr}= 0.2 \pi r- \frac{64}{r^2}[/tex].
    Notice the negative sign! The derivative of r-1 is -r-2.

    You should have [tex]r^3= \frac{0.2 \pi}{32}[/tex].
     
  4. Jun 28, 2006 #3
    oh i sorta get it now, well at least more than before, thanks
     
  5. Jun 28, 2006 #4
    thanks for the reply, however im not sure how u got the derivitive of the cost being equal to [tex]\frac{dC}{dr}= 0.2 \pi r- \frac{64}{r^2}[/tex]. Im not sure on how to achieve the value of 64 as i get a value of 32. My derrivitive: [tex]\frac{dC}{dr}= 0.2 \pi r- \frac{32}{r^2}[/tex]. Could u please explain how this is achieved? help greatly appriciated
    Pavadrin
     
  6. Jun 28, 2006 #5
    Yes, I think that was a typo from Halls .
     
  7. Jun 28, 2006 #6

    HallsofIvy

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    No, just stupidity!
     
  8. Jun 28, 2006 #7
    okay, fair enough, we all make mistakes sooner or later
    thanks anyhow
     
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